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At momentum transfers small compared with $M_W$, the two weak charged current vertices each carry $g/(2\sqrt{2})$ when written with $(1-{\gamma }_5)$, and the W boson propagator reduces to a local factor proportional to $1/M_W^2$. Matching their product to the Fermi interaction gives, at tree level,The Fermi constant is thus determined by the electroweak coupling and symmetry-breaking scale; the four-fermion description is a low-energy effective interaction.

Only the cross term $J_L^{†}{J}_H$ can annihilate the pion and create the stated charged-lepton/antineutrino pair. Its leptonic matrix element is ${\bar{u}}_e(k){\gamma }^{\mu }(1-{\gamma }_5)v_{\nu }(q)$. Lorentz covariance permits a vacuum-to-spin-zero current matrix element only proportional to $p^{\mu }$. At rest this has only a temporal component. The vacuum is parity-even and the pion is parity-odd, while $V_H^0$ is parity-even, so $⟨0|V_H^0|\pi ⟩$ equals its own negative and vanishes. Lorentz covariance then makes the full vector matrix element vanish. The temporal axial current is parity-odd and is allowed: this is the vacuum-to-pseudoscalar current selection rule.

The minus sign in $J_H=V_H-A_H$ cancels the interaction's minus sign, up to an irrelevant common S-matrix phase. ThereforeThe pion decay constant convention here includes the given $\sqrt{2}$ in its current matrix element. Momentum conservation and the external Dirac equations give $p=k+q$, ${\bar{u}}_e/k=m_e{\bar{u}}_e$, $/q{v}_{\nu }=0$. Since $/q(1-{\gamma }_5)=(1+{\gamma }_5)/q$, the neutrino term vanishes andThis is helicity suppression: the amplitude vanishes in the zero charged-lepton-mass limit despite the larger available phase space.

For the spin sum, the Dirac adjoint of $1-{\gamma }_5$ is $1+{\gamma }_5$, and the completeness relations giveThe chiral projectors automatically eliminate the unused neutrino helicity when the full massless completeness relation is inserted. There is no initial spin average for a spin-zero pion.

In the pion rest frame, let $r=|\mathbf{k}|=|\mathbf{q}|$, $E=\sqrt{r^2+m_e^2}$. After integrating the spatial delta function and angles, the two-body Lorentz-invariant phase space becomesThe root is $r_{\ast }=(m_{\pi }^2-m_e^2)/(2m_{\pi })$ and $E_{\ast }=(m_{\pi }^2+m_e^2)/(2m_{\pi })$. The absolute derivative of the delta-function argument is $r_{\ast }/E_{\ast }+1=m_{\pi }/E_{\ast }$. Thus ${\Phi }_2=r_{\ast }/(4\pi m_{\pi })$. Multiplying by the spin sum and the initial factor $1/(2m_{\pi })$ yields the leptonic pseudoscalar decay widthIf the decay constant were instead defined by a matrix element $i{f}_{\pi }{p}^{\mu }$ without $\sqrt{2}$, then $f_{\pi }=\sqrt{2}{F}_{\pi }$ and the width coefficient would be $f_{\pi }^2/(8\pi )$; mixing these conventions would introduce a factor-of-two error.

The same derivation for a muon cancels the common hadronic and weak factors in the pion electron-to-muon decay ratio:The numerical value uses approximate masses $0.511,105.66,139.57$ MeV. Agreement with the corresponding Standard Model prediction tests lepton universality and the chiral, helicity-suppressing structure of the weak charged current, rather than the poorly calculable hadronic decay constant. A difference in the electron and muon weak couplings would multiply the ratio by their squared coupling ratio. Precision agreement must be assessed after radiative corrections, with matching photon-inclusive experimental definitions; the displayed tree-level ratio is not itself the radiatively corrected precision prediction.

The product of leptonic weak charged currents in the Fermi interaction contains four-fermion cross terms such asThis produces ${\mu }^{-}\to e^{-}+{\bar{\nu }}_e+{\nu }_{\mu }$, and the same interaction permits neutrino-electron scattering and inverse muon decay. Pure electromagnetic interactions have no neutrino coupling and preserve the separate charged-lepton species, so cannot produce these processes by themselves. The factor $1-{\gamma }_5=2P_L$ fixes the chiral structure of the weak charged current.

The pion is a pseudoscalar with negative parity, while the strong-interaction vacuum is parity even. Lorentz covariance restricts a vacuum-to-pion vector-current matrix element to $a{p}^{\alpha }$. In the pion rest frame only its temporal component could be nonzero, but a vector current's temporal component is parity even, whereas the pion is parity odd. The matrix element therefore equals its negative and vanishes. The temporal component of an axial current is parity odd, so its matrix element is allowed and Lorentz covariance fixes it to a constant times $p^{\alpha }$. Defining the pion decay constant in the printed normalization givesThis proves both the vector-current exclusion and the axial-current form; $F_{\pi }$ itself is a strong-interaction parameter, not fixed by symmetry.

The cross term $J_{\mathrm{leptons}}^{\alpha †}{J}_{\alpha }^{\mathrm{hadrons}}$ gives, up to an irrelevant overall phase, the decay amplitudeTake the antineutrino mass to be zero. Since $p=k+q$, ${\bar{u}}_e/k=m_e{\bar{u}}_e$, and $/q{v}_{\nu }=0$, anticommutation with ${\gamma }_5$ givesThe amplitude has helicity suppression by one power of the charged-lepton mass.

Using the external spin sums, with no initial spin average for a spin-zero pion,The middle equality follows from $(1-{\gamma }_5)/q=/q(1+{\gamma }_5)$, $(1+{\gamma }_5{)}^2=2(1+{\gamma }_5)$, and the gamma matrix trace identities. Summing a full massless neutrino spin basis introduces no factor of two: the current projector annihilates the inactive helicity.

For completeness, evaluate the two-body Lorentz-invariant phase space in the pion rest frame. Its three-momentum delta function sets $\mathbf{q}=-\mathbf{k}$, and the energy delta function is $\delta (M_{\pi }-\sqrt{|\mathbf{k}{|}^2+m_e^2}-|\mathbf{k}|)$. The radial root isCombining its derivative with $1/(4E_k|\mathbf{k}|)$ and integrating solid angle givesThe invariant squared amplitude is independent of direction, so the supplied decay formula yields the leptonic pseudoscalar decay widthThe denominator is $4\pi$ because the matrix element contains $\sqrt{2}{F}_{\pi }$. With the alternative definition $f_{\pi }=\sqrt{2}{F}_{\pi }$, the same width is $G_F^2{f}_{\pi }^2{M}_{\pi }{m}_e^2(1-m_e^2/M_{\pi }^2{)}^2/(8\pi )$. In either convention it vanishes as $m_e\to 0$.

Replacing the electron by a muon gives the same hadronic factor and weak coupling. Hence the pion electron-to-muon decay ratio isThis is a tree-level result with massless neutrinos. The cancellation of $F_{\pi }$ and $G_F$ makes the ratio a clean test of the predicted helicity suppression and electron–muon universality in the chiral weak charged current. For example, scalar or pseudoscalar four-fermion interactions could remove the charged-lepton mass factor and substantially enhance the electron mode. Radiative corrections are required for a precision experimental comparison.

With two quark generations, the charged hadronic current containsA $K^{-}$ contains $s\bar{u}$, so its annihilation into the lepton pair selects the term proportional to $\sin {\theta }_C$. Strong strangeness conservation excludes the $d$ term from its vacuum matrix element. Its amplitude is thus multiplied by $\sin {\theta }_C$ and its width by ${\sin }^2{\theta }_C$:Here $F_K$ is the pure strong-interaction decay constant in the same square-root-of-two convention. If the pion current's quark mixing coefficient is also shown explicitly, its width carries ${\cos }^2{\theta }_C$; it is already absorbed if the stated pion matrix element uses the full weak current. Thus the isolated weak-factor ratio is ${\tan }^2{\theta }_C$, while the kaon suppression relative to unit weak coupling is the requested Cabibbo suppression factor ${\sin }^2{\theta }_C$.

The weak charged current creates a negative charged lepton in a left-chiral state. For a massless particle, that would have negative helicity; part (e) instead requires positive helicity. A nonzero mass supplies the necessary wrong-helicity component. This helicity suppression is extremely strong for the fast electron, but much weaker for the slow muon.

For a positive-helicity charged lepton with spinors normalized to unit probability, the left-chiral weight is $(1-{\beta }_{\ell })/2$. With the box-state normalization of part (c), this gives the useful consistent rate ratioIndeed $1-{\beta }_{\ell }=2m_{\ell }^2/(m_{\pi }^2+m_{\ell }^2)$ cancels the extra energy-dependent factor in that density ratio, leaving the standard pion electron-to-muon decay ratio in inverted form:Using the rounded masses in this paper gives $R\simeq 8185$, close to the stated order of $8000$. Precision predictions require more accurate masses and electromagnetic radiative corrections. The essential mechanism is the charged-lepton-mass dependence of the decay amplitude; simply counting available final momenta misses it.

The charged pion has spin zero. In its rest frame, the outgoing muon or Positron and the neutrino travel in opposite directions. With a nearly massless left-helicity neutrino, total angular momentum zero requires the positive charged antilepton to have negative helicity. The weak charged current instead creates a massless positive antilepton with positive helicity. A nonzero charged-lepton mass supplies the required opposite-helicity component.

Thus the decay amplitude is proportional to $m_{\ell }$: this is helicity suppression. Explicitly, contracting the pion's axial-current matrix element, proportional to its four-momentum $p_{\pi }^{\mu }$, with the leptonic current and applying the Dirac equation leaves a factor $m_{\ell }$. The leptonic pseudoscalar decay width is, in the $f_{\pi }$ convention,The phase space advantage of the lighter Positron cannot offset its much stronger helicity suppression. The common factors cancel in the pion electron-to-muon decay ratio:This tree-level estimate explains why the muonic branch dominates by roughly four orders of magnitude. The paper's table gives an electronic branching fraction $1.230\times 10^{-4}$; electromagnetic corrections change the simple ratio by a few percent without changing the mechanism.